Re: Worldmap mercator projection - Latitude to Y calculation.

"J. P. Gilliver" <[email protected]> Thu, 15 Jan 2026 17:25:26 +0000
Newsgroups alt.comp.os.windows-xp,alt.windows7.general,alt.comp.os.windows-10
Organization 255 software
Message-ID <[email protected]>
On 2026/1/14 8:24:17, R.Wieser wrote:
> Hello all,
> 
> I have a world map using the Mercator projection, and would like to plot 
> some stuff on it using Latitude and Longitude.
> 
> The problem is that I can't seem to get the formule I found to spit out the
> correct value for Y.
> 
> -- The map :
> 
> https://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Mercator_projection_Square.JPG/250px-Mercator_projection_Square.JPG
> 
> -- The formula :
> 
> Y = ln( tan(latitude) + sec(latitude) )
> 
> where "ln(...)" is log(...) / log(10)
> 
> ... at least, that is what I could google about it.

Rather than Googling the formula, my first thought was to go back to
first principles: as the name implies, a map projection can be thought
of as being created by shining a light through the earth (globe) onto a
sheet of paper - either flat and fixed to the globe at one point, or -
more commonly - a cylinder wrapped round the globe, touching at one
circle (often the equator), and then unrolled. Once this is realised,
basic geometry should make calculation of the co-ordinates fairly simple.

Unfortunately, to do this, one needs to know where the nominal light
source is. And - despite it being a _long_ article - I can't find a
statement of this in the Wikipedia article about the Mercator projection
- other that that - I _think_ - it _isn't_ the centre of the earth
(so-called "radial"), though is often thought to be (certainly Google's
AI thinks it is).

If anyone _can_ find out where the light source point is for the
Mercator projection, I'd love to know! (And it would answer Rudy's
question.)

(One other position for the light source I remember from last time I
looked into this - which was probably over 40 years ago! - is on the
opposite surface of the globe to the projection point (i. e. sort of
tracking round opposite the map "printing"); I don't think it's that,
though, as that would show the poles, though still distorted.

If it _is_ light-source-at-centre, then the Y co-ordinate would just be
the tangent of the latitude (scaled appropriately for the map size).
Even if it isn't, this _may_ be close enough - try a few places.

-- 
J. P. Gilliver. UMRA: 1960/<1985 MB++G()ALIS-Ch++(p)Ar++T+H+Sh0!:`)DNAf

It has been my experience that folks who have no vices have very few
virtues
-- Abraham Lincoln, quoted by Mark Lloyd in alt.windows7.general 2018-12-27